To evaluate the integral ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{e^{6x}}{e^{3x}+13}dx$ the most appropriate substitution is $u=$ in which case $du=$

This allows the integral to be expressed as ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}f(u)du,$ with $f(u)=$ Since ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}f(u)du=$ we obtain

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{e^{6x}}{e^{3x}+13}dx=$

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